A long straight wire AB carries current 10 A. A proton (P) travels with a speed $2×10^6 m/s$, parallel to the wire, at a distance 2 m from it in a direction opposite to the current as shown in figure. The force experienced by the proton is: $(μ_0=4π×10^{-7}NA^{-2})$
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → $3.2×10^{-19}N$
The magnetic field (B) at a distance (r) from a straight wire carrying current (I) is:
$B=\frac{μ_0I}{2πr}$ [formula]
$μ_0=4π×10^{-7}$
$I=10A$
$r=2m$
$B=\frac{4π×10^{-7}×10}{2π×2}=1×10^{-6}T$
Now,
Magnetic force, $F=qvB\sin θ$
where,
q, charge of proton = $1.6×10^{-19}C$
v, velocity of proton = $2×10^6m/s$
B, Magnetic field = $1×10^{-6}T$
θ, ($\vec v$ is perpendicular to $\vec B$) = 90°
$F=1.6×10^{-19}×2×10^6×1×10^{-6}×\sin 90°$
$=3.2×10^{-19}N$